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Leetcode #1585: Check If String Is Transformable With Substring Sort Operations

In this guide, we solve Leetcode #1585 Check If String Is Transformable With Substring Sort Operations in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given two strings s and t, transform string s into string t using the following operation any number of times: Choose a non-empty substring in s and sort it in place so the characters are in ascending order. For example, applying the operation on the underlined substring in "14234" results in "12344".

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Greedy, String, Sorting

Intuition

A locally optimal choice leads to a globally optimal result for this structure.

That means we can commit to decisions as we scan without backtracking.

Approach

Sort or preprocess if needed, then repeatedly take the best available local choice.

Maintain the minimal state necessary to validate the greedy decision.

Steps:

  • Sort or preprocess as needed.
  • Iterate and pick the best local option.
  • Track the current solution.

Example

Input: s = "84532", t = "34852" Output: true Explanation: You can transform s into t using the following sort operations: "84532" (from index 2 to 3) -> "84352" "84352" (from index 0 to 2) -> "34852"

Python Solution

class Solution: def isTransformable(self, s: str, t: str) -> bool: pos = defaultdict(deque) for i, c in enumerate(s): pos[int(c)].append(i) for c in t: x = int(c) if not pos[x] or any(pos[i] and pos[i][0] < pos[x][0] for i in range(x)): return False pos[x].popleft() return True

Complexity

The time complexity is O(n×C)O(n \times C)O(n×C), and the space complexity is O(n)O(n)O(n). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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